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Caller
Poland8075 Posts
I have a Calculus midterm tomorrow and I have no idea what to do for an n-epsilon proof. Delta epsilon isn't so bad right now, but n-epsilon is. From what I understand, according the the precise definition of a limit, lim x->infinity of a sequence a_n --> L and I have to set it so absolute value |a_n-L| < epsilon or something like that. And I have no idea what the hell this means. Halp anybody???  | ||
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Kwidowmaker
Canada978 Posts
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Muirhead
United States556 Posts
if and only if For any positive real number epsilon, there exists a positive integer N such that |a_m-L|<epsilon whenever m>N. | ||
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Caller
Poland8075 Posts
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zeks
Canada1068 Posts
i agree with muirhead's proof, but you might want to check your notes/textbook because you might be used to another set of notation | ||
micronesia
United States24776 Posts
On October 20 2008 06:03 zeks wrote: write it out 10 times and then just regurgitate it during the midterm =/ i agree with muirhead's proof, but you might want to check your notes/textbook because you might be used to another set of notation A good professor will prevent you from succeeding on a question about this simply by regurgitating the exact proof. One possible way is your second statement. Also, theoretical questions that require a moderate understanding of the principles behind the proof are another way the professor can force students to truly learn it. | ||
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Kwidowmaker
Canada978 Posts
Provide a converging sequence, a_n, and we can show a proof for an example. A monotone decreasing sequence is easiest. | ||
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fight_or_flight
United States3988 Posts
On October 20 2008 05:00 Caller wrote: Hi everybody I have a Calculus midterm tomorrow and I have no idea what to do for an n-epsilon proof. Delta epsilon isn't so bad right now, but n-epsilon is. From what I understand, according the the precise definition of a limit, lim x->infinity of a sequence a_n --> L and I have to set it so absolute value |a_n-L| < epsilon or something like that. And I have no idea what the hell this means. Halp anybody??? Since epsilon is a constant number, and |a_n-L| should always be getting smaller (because its converging), you should always be able to reach a point where |a_n-L| < epsilon for some (probably large) n. | ||
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Wonders
Australia753 Posts
So to formalize this, you let epsilon be any number, which is as small as you want it to be (but not zero). Then if you go far enough along the sequence, then the difference between terms in the sequence and the limit are going to be smaller than epsilon (which is as small as you like). That is, there exists some possibly huge number M, such that all terms in the sequence after M are going to be closer to the limit than epsilon, or |a_n - L| < epsilon for all n>M. | ||
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